# Is the Kernel a Maximal Ideal?

In this MSE post, Artur Araujo gives an alternative proof that the preimage of a maximal ideal under a surjection is a maximal ideal:

If I may suggest, a cleaner way of proving this is by an altogether different method, bypassing elements. Since $M$ is maximal, $k:=S/M$ is a field. Since $f$ is surjective, its composition with the canonical projection $\bar{f}:R\to k$ is a surjection. This means that $\ker \bar{f}$ is a maximal ideal. Can you compute it?

I am being dumb, so I don't see how $\ker \overline{f}$ is a maximal ideal in $R$. Here is a comment I left on Artur Araujo's post:

Okay. Let $f:R→F$ be a surjective ring homomorphism, where $R$ is some (commutative?) ring and $F$ is a field. But suppose that $\ker f$ is not maximal. Then there exists an ideal $I$ in $R$ such that $\ker f \subset I \subset R$ and therefore $\{0\} \subseteq f(I) \subseteq S$. This would be a contradiction if images generally preserved strict inclusions, but I believe this is only true of injective maps. So I don't see why $\ker f$ has to be maximal in $R$.

What's going on? For instance, if $f(x) = x^2$ over the reals, then $(-1,0] \subset (-1,1)$ but $f((-1,0]) = [0,1) = f((-1,1))$, which is equality, not strict inclusion. Do images preserve strict set inclusion in this case?

In a commutative ring $R$, an ideal $I$ is maximal iff $R/I$ is a field. Since $\overline{f}: R \to k$ is surjective, then $R/\ker(\overline{f}) \cong k$ by the First Isomorphism Theorem.
I don't see how $\ker \overline{f}$ is a maximal ideal in $R$.
It is maximal since $R/\ker(\overline f)$ is isomorphic to a simple ring. A simple ring only has trivial ideals, so the quotient only has two ideals.
The map between $R/\ker(\overline f)$ and it’s range is injective — it’s an isomorphism. And consequently, the ideals are in correspondence , and the correspondence even respects inclusion.
$f(x)=x^2$ is not a ring homomorphism, and I’m not even sure you’re considering any ring structure at all on the function’s domain you chose. At any rate, I don’t think you need to pursue that line of thought. If you invest a bit of thought in the answer you are referencing you’ll have a clear answer.